The heights of young men follow a Normal distribution with mean $\mu M=69.3$inches and standard deviation $\sigma M=2.8$inches. The heights of young women follow a Normal distribution with mean $\mu W=64.5$inches and standard deviation $\sigma W=2.5$inches. Suppose we select independent SRSs of $16$young men and $9$young women and calculate the sample mean heights $\overline{x}Mand\overline{x}W$.

a. What is the shape of the sampling distribution of $\overline{x}M-\overline{x}W$? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

It is given that there are two distributions that is,

$$\begin{gathered} DistributionM:Normalwith{\mu }_M=69.3,{\sigma }_M=2.8 \\ DistributionW:Normalwith{\mu }_W=64.5,{\sigma }_W=2.5 \end{gathered}$$

Now, if M and B are normally distributed then there difference$M-B$is also normally distributed.

And we know that the properties of normal are:

$$\begin{gathered} {\mu }_{aX+bY}=a{\mu }_x+b{\mu }_Y \\ {\sigma }_{aX}+{\sigma }_{bY}=\sqrt{a^2{{\sigma }^2}_x+b^2{{\sigma }^2}_y} \end{gathered}$$

Then we obtain:

$$\begin{gathered} {\mu }_{M-W}={\mu }_M-{\mu }_W \\ =69.3-64.5 \\ =4.8 \\ {\sigma }_{M-W}=\sqrt{{{\sigma }^2}_M+{{\sigma }^2}_W} \\ =\sqrt{2.8^2+2.5^2} \\ =3.7537 \end{gathered}$$

Thus, we conclude that the distribution of $M-W$is normal with mean and standard deviation as:

$$\begin{gathered} {\mu }_{M-B}=4.8 \\ {\sigma }_{M-B}=3.7537 \end{gathered}$$

$$\begin{gathered} {\mu }_1=69.3 \\ {\mu }_2=64.5 \\ {n}_1=16 \\ {n}_2=9 \\ {\sigma }_1=2.8 \\ {\sigma }_2=2.5 \end{gathered}$$

The mean of the sampling distribution of the difference in sample means is the difference in the population means. This implies:

$$\begin{gathered} {\mu }_{x_i-x_2}={\mu }_1-{\mu }_2 \\ =69.3-64.5 \\ =4.8 \end{gathered}$$

Thus, we conclude that the mean of the sampling mean is$4.8$inches.

Since the sample $16$young men is less than $10\%$of all young men and since the sample of $9$young women is less than $10\%$of all young women. Thus, the standard deviation is as follows:

$$\begin{gathered} {\sigma }_{x_1-x_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}} \\ =\sqrt{\frac{2.8^2}{16}+\frac{2.5^2}{9}} \\ =1.0883 \end{gathered}$$

Thus, we conclude that the difference in the sample means are expected to be vary by$1.0883$ inches from the mean difference of$4.8inches$.